The Berkeley Review MCAT Biology Part 2

540 Pages·2011·78.91 MB·English

by  The Berkeley Review

Biology
IV Partnl
Molecular Cell Biology
Sections YI-X
Section VI
Structure and Function
in Cells and Viruses
Section VII
Metabolic Components
Section VIII
Metabolic Pathways
Section IX
Genetic Information
Section X
Expression of
Genetic Information
The
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Biology
A. Biological Molecules
Section VI 1. Amino Acids and Proteins
2. Carbohydrates and Lipids
3. Nucleic Acids
Structure and
B. Eukaryotic Cells
Function in 1. Cellular Reproduction
a. Eukaryotic Chromosomes
Cells and Viruses
b. The Cell Cycle, Mitosis, and Meiosis
2. Cellular and Molecular Organization
a. Biomembranes and Membrane Transport
b. Nucleus, Nucleolus, and Ribosomes
c. Endoplasmic Reticulum, Golgi Apparatus,
Lysosomes, Peroxisomes, and Mitochondria
d. Microtubules, Microfilaments,
and Intermediate Filaments
C. Prokaryotic Cells
1. Cellular Reproduction
a. Prokaryotic Chromosomes
b. The Cell Cycle
2. Cellular and Molecular Organization
a. Biomembranes and Membrane Transport
b. Nucleoid and Ribosomes
D. Viruses
1. Architecture and Genomes
2. Infection and Genomic Expression
3. Assembly and Release of Progeny Virons
Practice Passages and Answers
T/te
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Structure and Function in Cells and Viruses
Top 10 Section Goals
Be familiar with the basic biological molecules of cells and viruses.
The four general classes ofbiological molecules are amino acids, nucleic acids, carbohydrates, and
lipids. Be able to recognize their basic differences, and know the functional significance ofeach class.
Know the twenty standard amino acids that are used to synthesize proteins.
Beable to associate amino acid names with their structures,especially the five amino adds (Asp,
Glu, Lys, Arg, andHis) thatcanhave charged polar side chains.
Understand the difference between mitosis and meiosis at the cellular level.
Know the different stages ofthecell cycle andhow they relate to one another. Be able to describe
P the chromosome numberaridploidylevel at anystageofmitosis or meiosis.
Be familiar with the different types of membrane transport processes.
Have a feel for the differences betweensimplediffusion, facilitated diffusion, active transport, and
bulk transport Know something about symports, antiports, uniports, andmembrane charge balance.
Be familiar with the structure of membrane-bound organelles in a eukaryotic cell.
Be awareofhow theyrelate to oneanother within theircellular environment. Know theirgeneral
* structuralcharacteristics (e.g., single membrane, double membrane, contains DNA, etc.).
Be familiar with the functions of eukaryotic organelles.
Foreachof the majororganelles in a eukaryotic cell, be ableto define its function in generalterms.
Is it involved in metabolism? Cellular packaging and sorting? Replication? Proteinsynthesis?
Know where the different metabolic processes occur in a cell.
The general metabolic reactions of glycolysis, the Krebs cycle, electrontransport, and oxidative
•> phosphorylation are common toeukaryotes andprokaryotes. Where do they occur ineach cell?
Know the general differences between Gram-negative and Gram-positive bacteria.
Bacterial cells, for the most part, canbe divided into Gram-negative and Gram-positive cells. It is
important tohavean understanding ofthestructural differences between the twocell types.
Be familiar with the reproductive cycle of a virus.
Thebiological diversity withinviruses is enormous. Justhavea general idea of the strategies used
byenveloped andnonenveloped viruses when they infect ahost cell.
Know the general differences between eukaryotes, prokaryotes, and viruses.
It is important to knowthebasic differences between eukaryotic cells, prokaryotic cells, and viruses.
Havean understanding ofthe size differences. Howdo theyinteract with eachother?
Biology Structure &? Function in Cells &Viruses Biological Molecules
Biological Molecules
Amino Acids & Proteins™
Amino Acids
Even though there are more than 400 naturally occurring amino acids, there are a
standard set of 20 amino acids that comprise the proteins of all living species. As
we will see, proteins are simply individual amino acid residues linked together
in a head-to-tail fashion. All of the standard 20 amino acids are referred to as o>
aminoacids (i.e., a 2-amino acid), except forproline which is referred to as an oc-
imino acid.
An amino acid is composed ofa basic amino group (-NH2), an acidic carboxyl
group (-COOH),a hydrogen (H) atom, and a side chain which is characteristic to
each amino acid. If no one particular amino acid is being discussed, the side © coo
chain is usually designated as -R. The amino group of an amino acid is covalently H,N- C- H
IOC
attached to the a-carbon atom of the amino acid-hence the name oc-amino acid.
The a-carbon atom (Figure 6-1) is the carbon atom next to the carboxyl carbon
atom. Figure 6-1
An oc-amino acid.
Each of the 20 standard amino acids has a particular three letter and one letter
abbreviation. For example, proline can be abbreviated either as Pro or as P. Each
amino acid also has a defined molecular weight (Mr) and can be placed,
according to its R group (at pH 7.0), into one of three main families of amino
acids. Thefirst family (Figure6-2) consists of those amino acids with nonpolar R
groups. These amino acids are hydrophobic.
Nonpolar R Groups
© HI OII © © Hl OII © © Hl Oll © © Hl Oll © © H1 O11 ©
H3N— C- C- O H,N— C
I — C- O H3N— CI — C- O H3N— CI — C- O H3N— CI - C- O
H CH3 H3C— CH CH, H3C— CH
I l I
Glycine (Gly)[G] CH, H3C— CH CH2
Mr = 75 Alan
Miner=(A8l9a)[A] CI H3 CI H3
Valine (Val)[V] Leucine (Leu)[L] Isoleucine (Ile)[I]
Mr=117 Mr=131 Mr=131
H O H O
© I ll 0 © 1 11 © © 0 ©
H2N— C- C- O H3N— C— C- O H,N—C-C- O H3N—C- C- O
i I
u CI H,
CH2
Pro
Mlinre=(1P1ro5)[P] CSI H3
Methionine (Met)[M] Phenylalanine (Phe)[F] Tryptophan (Trp)[W]
Mr= 149 Mr=165 M. = 204
Figure 6-2
Amino acids with nonpolar side chains.
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Biology Structure &Function in Cells 6t Viruses Biological Molecules
Thesecondfamily consists of those aminoacidswith polar uncharged R groups
(Figure 6-3). These amino acids are hydrophilic because they contain side chain
functional groups(e.g., -OH, -SH, or -NH2) that canhydrogenbond with water,
Polar Uncharged R Groups
H O H H O
© l 11 © I © © l 11
H3N— C— C- O H3N— C— C- O H3N— C— C- O
3 I 3 I I
CH2 HO-CH CH,
I
OH CH3 SH
Serine (Ser)[S] Threonine (Thr)[TJ Cysteine (Cys)[C]
Mr=105 Mr=119 M,= 121
H O ~ H o H O
© l 11 © I ll © © l ll ©
H,N— C- C- O H3N— C- C- O H3N- C— C~ O
I J I
CH2 CH2
I I
0= C- NH2 CH2
l
0= C- NH2
Asparagine (Asn)[N] Glutamine (Gln)[Q]
Mr=132 Mr=146
Tyrosine (Tyr)[Y]
M,= 18I
Figure 6-3
Amino acids with uncharged polar side chains.
The third family (Figure 6-4) consists of those amino acids with charged R
groups. Note that two of the amino acids (Asp and Glu) have side chains which
are carboxylic acids and three of the amino acids (Lys, Arg, and His) have side
chains which are amine bases. The side chains of Asp, Glu, Lys, and Arg are
highly ionized at neutral pH. The side chain of His is weakly ionized at neutral
pH. We will see that the degree of ionization of an amino acid depends on its
acidic and basic properties.
Charged R Groups H 0 H O
© l ll © © l 11 ©
H3N— C— C--O H3N-• c— c- 0
3 1 1
H 0
H 0 CH2 CH2 © I ll ©
© 1 ll © 1 1 H3N— C— C- 0
H 0 H3N— C— C- 0 CH2 CH2 3 1
© l II © 1 1 1 CH2
H3N— C— C- 0 CH2 CH2 CH2
3 1 1 1
CH2 NH r^©
1 2© 1 2© ©' l © H-N^ .N-H
0=C—0 0=C—0 NH3 H2N-• C= NH2 N^
Aspartate (Asp)[D] Glutamate (Glu)[E] Lysine (Lys)[K] Arginine (Arg)[R] Histidine (His)[H]
Mr=133 Mr=147 Mr=146 M r=174 Mr=155
Figure 6-4
Amino acids with charged side chains.
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Although the standard 20 amino acids can exist in either the D or the L
configuration (with the exception of glycine because it is non-chiral), only the L-
amino acidsare found in proteins. It is not knownwhy evolution chose to incor © ©
coo coo
porate L-amino acids into proteins instead ofD-amino acids. As anexample, the
© 7 ©
two non-superimposable mirror images of the a-amino acid alanine are shown in H'.I1,N»-C-* H Mi C-^ NH
Figure 6-5. Note that the a-carbon of alanine is chiral (there are 4 different
CH, CH,
substituents attached to that a-carbon). The D isomer and the L isomer of alanine Mirror
are called enantiomers. The D and L isomers of the different amino acids are
L-alanine D-alanine
based on the absolute configuration ofD and L-glyceraldehyde. [All amino acids
mentioned in these readings will be in the L-configuration, unless otherwise Figure 6-5
stated.] Non-supcrimposable mirror images.
Even though D-amino acids are never found in proteins, they do exist in many
organisms. For example, D-alanine and D-glutamate are found within the rigid
cell walls ofsome bacteria. The presence ofthese D-amino acids helps to prevent
the degradation of the bacterial cell wall by specific enzymes called proteases
(more on this later). D-Valine is found in the antibiotic valinomycin, a carrier of
K+ ions across membranes.
Amino acids found within proteins can be modified as shown by the examples
given in Figure 6-6. Two important amino acid modifications can be found in
collagen, the most abundant protein inmammals. Some ofthe proline and lysine
amino acid residues found within nascent molecules of collagen are
hydroxylated to give 4-hydroxyproline and 5-hydroxylysine, respectively. These
modified aminoacidshelp togivecollagen itshigh tensile strength. Histamine, a
powerful vasodilator, is formed from the decarboxylation of the amino acid
histidine. S-Adenosylmethionine is an activated form of the amino acid
methionine and can methylate protein or nucleic acid substrates.
« H ° NH,
© I II ©
~ H ° H^N—C-C-0
© i II 0 I
H3N— C— C- O H CH,
© l
I '
CH, !i.--:-.-i! CH,
i
CH, ©S-
0 H-C— OH CH,
C-0 © I H-Nv . N
H3N— CH2
HO OH
4-Hydroxyproline 5-Hydroxylysinc Histamine S-Adenosylmethionine
Figure 6-6
Modified amino acids.
Acid-Base Properties
Body fluids in biological systems generally have a pH range of 6.5 to 8.0. This is
considered to be physiological pH. At these pH ranges in the cell the amino and
carboxyl groups of the standard amino acids are ionized. In other words, the a-
amino group bears a positive charge while the a-carboxyl group bears a negative
charge. The dipolar or zwitterionic nature of the amino acids is due to the fact
that the pKa value for the a-amino terminal is about 9.4 while the pKa for the a-
carboxyl terminal is about 2.2. Since amino acids can act as either an acid or a
base they are referred to as ampholytes. Table 6-1 shows the pKa values for the
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a-carboxyl and a-amino groups of a selectionof the 20. standard amino acids and
their ionizable side chains. Small pKa values refer to a strong acid while large pKa
values refer to a weak acid.
Amino Acid pK^a-COOH) pKa(a-NH3+) pl^ (side chain groups)
Glycine 2.4 9.8 —
Alanine 2.3 9.9 —
Aspartic Acid 2.0 9.9 3.9 (P-COOH)
Glutamic Acid 2.2 9.7 4.2 (Y-COOH)
Histidine 1.8 9.2 6.0 Imidazole)
Cysteine 1.8 10.8 8.3 (Sulfhydryl)
Tyrosine 2.2 9.1 10.1 (Phenol)
Lysine 2.2 9.2 10.8 (e-amino)
Arginine 1.8 9.0 12.5 (Guanidino)
Serine 2.2 9.2 13.0 (Hydroxyl)
Table 6-1
The pK values for the ionizinggroups of a few of the standard 20 amino acids.
The Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation (6-1) can be used to determine the fraction
of either the a-amino, a-carboxyl, or side chain groups that are ionized at a
particular pH. In thisequation [Ae] is the concentration of conjugate basewhile
[HAj is the concentration of conjugate acid.
[A']
pH = pKa + log
(6-1) [HA] Henderson-Hasselbalch Equation
Let's consider the ratio of the protonated to the unprotonated a-amino group of a
general amino acid at a pH of 7.0. The pKa for a typical a-amino group is about
9.4. Using the Henderson-Hasselbalch equation, we find that the ratio of the
protonated a-amino group to the unprotonated a-amino group is about 102-4 or
251:1.This tells us that the predominant form of the a-amino group at a pH of 7.0
pH-PKa [A"]/HA
is the protonated form. The same type of analysis for the a-carboxyl group tells
-3 0.001 us that the ratio of the a-carboxylate anion to the protonated a-carboxyl group is
-2 0.01 about 63,000:1.
-1 0.1
0
1 101 (6-2) 7.0 = 9.4 + log [R-NH2] 2.4 =[R-NH?]=25I (6_3)
[R-NHJ] [R-NH2] 1
2 100
3 1000
Similarly, at a neutral pH of 7.0 the ratio of the protonated to non-protonated
Table 6-2 side chain of histidine is about 1:10.
Difference between pH and
pKa and the relationship to the If we know the difference between the pH and the pKa (i.e., pH - pKa), then we
ratio of conjugate base to can establish a ratio (Table 6-2) between the concentration of conjugate base to
conjugate acid. conjugate acid. If a particular amino acid is required at the catalytic site of an
enzyme, knowing the ratio between conjugate base and conjugate acid will help
determine the maximum potential activity of the enzyme.
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Isoelectric Point (pi)
The isoelectric point (pi) is that p H at which an amino acid (or a molecule)
carries no net electric charge. By manipulating the Henderson-Hasselbalch
equation wecanobtain an expression (6-4) thatstates that the isoelectric point is
simply the arithmetic mean of two pKa values.
, _ [pKat + pKa2]
(6-4) P 2
For example, the isoelectric point for the amino acid glycine is 6.1 while the
isoelectric point for the amino acid lysine is 10.0. If we were to place these two
amino acids in an electric field, we would find that at any pH above their
isoelectric points they would migrate toward the anode (positive electrode) while
at any pH below their isoelectricpoints they would migrate toward the cathode
(negative electrode). Thischaracteristic allows forseparation ofamino acids by a
process called electrophoresis. Paper electrophoresis is generally used for
separating mixtures that contain charged molecules which are small while gel
electrophoresis is used for separating proteins and nucleic acids.
Titration Curves
At physiological pH glycineexists in solution as the dipolar ion. The net charge
on this amino acid would be 0. However, ifwe were to add a strong acid to the
dipolar solution, the carboxylate group would become protonated and the
overallcharge of the amino acid would be +1.Similarly, addition of a strong base
to the dipolar solution would remove a proton from the protonated a-amino
group, giving the amino acid a net charge of -1. As shown in Figure 6-7, the
ionization of glycine depends on the pH of the solution.
PH=1 pH « 6.1 pH«14
~ H O © ^ H ° © H O
© I II © i ii e I II 0
H,N- C— C- OH =5= H3N— C- C- O H2N— C— C- O
pKa H pKa H
2.4 9.8
[Net charge is+lj [Net charge is 0] [Net charge is-1]
Figure 6-7
Ionization of glycine.
The ionization of glycine can be followed by using a titration curve (Figure 6-8).
We can start our titration by adding equivalents of base (such as NaOH) to a
solution of glycine which is fully protonated. The initial pH of this solution might
be in the neighborhood of 1 and the overall net charge on glycine is roughly +1.0.
The first midpoint of the titration curve comes when [NH3®CH2COOH] =
[NH3®CH2COOe]. The Henderson-Hasselbalchequation tellsus that at this first
midpoint the pH = pKa. In other words, the pH at this point is 2.4 (which is the
pKa of the first ionizable proton), and half of the solution is composed of
NH3®CH2COOH while the other half of the solution is composed of
NH3®CH2COOe. The overall net charge on glycine in solution at this point is
about +0.5.
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Addition of more base will eventually convert all of the NH3®CH2COOH to
NH3®CH2COOe. At this point the first equivalence point of the solution has
been reached. The overall net charge on glycine is 0. The isoelectric point has
been reached and the pH is about 6.1.
In order to reach the second midpoint in the titration we add more base. At the
second midpoint wefind that [NH3®CH2COOe] = [NH2CH2COO®]. Again, the
Henderson-Hasselbalch equation tells us that at the second midpoint the pH =
pKa. The pH at this point is 9.8 and the overall net charge on glycine in solution
is about -0.5.
Second Midpoint
12n
[NH3+CH2COO-] = [NH2CH2COO-]
0.5 1.0
[NH3+CH2COOH] Equivalence ofNaOH Added
Figure 6-8
Titration of glycine.
By the time the second equivalence point is reached enough base has been added
to completely titrate all the ionizable protons on glycine. The dominant species in
solutionis now NH2CH2COOe. Theoverallnet charge on the amino acid is -1.
Buffers
The ability of a solution to resist a change in its pH when either an acid or a base
is added is the principle behind the buffer capacity of a solution. In Figure 6-8
note that the slope of the titration curve is much less near the first and second
midpoints than near the equivalence point. For each increment of base added in
the vicinity of the pKa's of glycine the pH changes very little. This means that
the numerical value of the log ([Ae]/[HA1) in the Henderson-Hasselbalch
equation doesnot change thatmuch. In otherwords, [Ae] approximately equals
[HA] in the region of the pKa. If a weak acid is within 1 pH unit of its pKa value,
it resides within a good buffering range.
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